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3 years ago

Let the random variable X denote the number of network blackouts in a day. The

1 answer:

6 0

E(X) = 0(0.7) + 1(0.2) + 2(0.1) = 0.2 + 0.2 = 0.4
The expected daily loss due to blackouts = 0.4 * $500 = $200

Var(X) = 0(0.7 - 0.4)^2 + 1(0.2 - 0.4)^2 + 2(0.1 - 0.4)^2 = 0.04 + 0.18 = 0.22
The expected daily variance due to blackouts = 0.22 * $500 = $110

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Which shape has four congruent sides and two sets of parallel sides?

Dovator [93]

It’s a Parallelogram

6 0

2 years ago

What is 832.248 rounded to the nearest tenth? A. 830 B. 832.2 C. 832.25 D. 832.3

OleMash [197]

Hi there.

To start, we have to understand where the ten the place is.

Hint: it's 1 place behind the decimal point.

Now we need to know what numbers behind the tenths place make it round up or down.

If the number behind it is 4 or under, round down.

5 & up, round up.

In this case, 830.248 has 2 numbers behind the decimal, so we have to round those THEN the 2 in the tenths place.

830.25 to 830.3

Your answer is D. 830.3

I know this was lengthy, sorry. Just wanted to explain :p

4 0

3 years ago

a park is 4 times as long as it is wide. if the distance around the park is 12.5 kilometers, what is the area of the park?

blsea [12.9K]

So, first, what are the two sides?

let's call then x and y

we know that 2(x+y)=12.5 (that's the distance around)

so that means that x+y=6.25 (I just divided both by 2)

now, x=4y (from "4 times as long as it is wide")

so we can substitute:

x+4x=6.25

5x=6.25

x=1.25

so one side, is 1.25 and the other will be 1.25*4=5

and for the area we multiply the two:

1.25*5=6.25 square kilometers, and this is the answer!

7 0

3 years ago

Read 2 more answers

Help please! worth a lot of points! Screenshot included

vlada-n [284]

Answer:

x=44444444444444444444444444444444

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)

S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

4 0

3 years ago